We have a two independent random variables X and Y with mu_x = 50 and sigma_x =

Question

We have a two independent random variables X and Y with mu_x = 50 and sigma_x = 5 and mu_Y = 60 and sigma_y - 10 If P(50 - a X 50 + b) = 0.815, find a and b What is the probability that random variable X falls between 35 and 45 Find, P[(35 X 45)and (45 Y 70)] A company that produces fine crystal knows from experience that 10% of its goblets have cosmetic flaws and must be classified as "seconds" P(X = x) = Q) * px * (1 - p)n~x you can use your TI-89; you need to figure out if this is a "combination" or "permutation" problem Among six randomly selected goblets, how likely is it that only one is a second? Among six randomly selected goblets, what is the probability that at least two are seconds? If goblet are examined one by one, what is the probability that at most five must be selected to find four that are not seconds

Explanation / Answer

10.

a)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.815      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.0925      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.3255162      
By symmetry,          
z2 =    1.3255162      
          
As          
          
sigma(X) = standard deviation =    5      
          
Then

a = z*sigma(X) = 1.3255162*5 = 6.627581 [ANSWER]

b has the same value by symmetry,

b = 6.627581 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    35      
x2 = upper bound =    45      
u = mean =    50      
          
s = standard deviation =    5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -3      
z2 = upper z score = (x2 - u) / s =    -1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.001349898      
P(z < z2) =    0.158655254      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.157305356   [ANSWER]

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c)

FOR Y:

We first get the z score for the two values. As z = (x - u) / s, then as          
Y1 = lower bound =    45      
Y2 = upper bound =    70      
u = mean =    60      
          
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (Y1 - u)/s =    -1.5      
z2 = upper z score = (Y2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.066807201      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(45 < Y < 70) =    0.7745375

Hence,

P[(35<x<45) and (45<y<70)] = 0.157305356*0.7745375 = 0.121838897 [ANSWER]

  

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